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Transcript
```1
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47
6
Journal of Integer Sequences, Vol. 17 (2014),
Article 14.2.8
23 11
Extremely Abundant Numbers
and the Riemann Hypothesis
Departamento de Matemática
4169-007 Porto
Portugal
[email protected]
[email protected]
Abstract
Robin’s theorem states that the Riemann hypothesis is equivalent to the inequality
σ(n) < eγ n log log n for all n > 5040, where σ(n) is the sum of divisors of n and γ is
Euler’s constant. It is natural to seek the first integer, if it exists, that violates this
inequality. We introduce the sequence of extremely abundant numbers, a subsequence
of superabundant numbers, where one might look for this first violating integer. The
Riemann hypothesis is true if and only if there are infinitely many extremely abundant
numbers. These numbers have some connection to the colossally abundant numbers.
We show the fragility of the Riemann hypothesis with respect to the terms of some
supersets of extremely abundant numbers.
1
Introduction
There are several statements equivalent to the famous Riemann hypothesis (RH) [4]. Some
of them are related to the asymptotic behavior of arithmetic functions. In particular, the
well-known Robin theorem (inequality, criterion, etc.) deals
P with the upper bound of the
sum-of-divisors function σ, which is defined by σ(n) :=
d|n d. Robin [24, Theorem 1]
1
established an elegant connection between RH and the sum of divisors of n by proving that
the RH is true if and only if
σ(n)
< eγ ,
n log log n
for all n > 5040,
(1)
where γ is Euler’s constant.
Throughout the paper, as in Robin [24], we define the function f by setting
f (n) =
σ(n)
.
n log log n
(2)
Gronwall, in his study of the asymptotic maximal size for the sum of divisors of n [11], found
that the order of σ(n) is always “very nearly n” [12, Theorem 323]. More precisely, he proved
the following theorem.
Theorem 1 (Gronwall). Let f be defined as in (2). Then
lim sup f (n) = eγ .
(3)
n→∞
Let us call a positive integer n [2, 23]
(i) colossally abundant, if for some ε > 0,
σ(m)
σ(n)
≥ 1+ε ,
1+ε
n
m
(m < n)
and
σ(n)
σ(m)
> 1+ε ,
1+ε
n
m
(m > n);
(4)
(ii) generalized superior highly composite, if there is a positive number ε such that
σ−s (m)
σ−s (n)
σ−s (m)
σ−s (n)
≥
,
(m < n)
and
>
,
(m > n),
ε
ε
ε
n
m
n
mε
P
where σ−s (n) = d|n d−s . The parameter s is assumed to be positive in [23]. In the
case s = 1, (ii) becomes (i).
Ramanujan initiated the study of these classes of numbers in an unpublished part of his 1915
work on highly composite numbers ([21, 23] and [22, pp. 78–129, 338–339]). More precisely,
he defined rather general classes of these numbers. For instance, he defined generalized highly
composite numbers, containing as a subset superabundant numbers [21, Section 59], and he
introduced the generalized superior highly composite numbers, including as a particular case
colossally abundant numbers. For more details we refer the reader to [2, 10, 23].
We denote by CA the set of all colossally abundant numbers. We also use CA as an
abbreviation for the term “colossally abundant”. Ramanujan [23] proved that if n is a
2
generalized superior highly composite number, i.e., a CA number, then under the RH we
have
p
√
σ(n)
γ
log n ≥ − eγ (2 2 + γ − log 4π) ≈ −1.558,
− e log log n
lim inf
n→∞
n
p
√
σ(n)
γ
lim sup
log n ≤ − eγ (2 2 − 4 − γ + log 4π) ≈ −1.393.
− e log log n
n
n→∞
Robin [24] also established (independent of the RH) the following inequality
f (n) ≤ eγ +
0.648214
,
(log log n)2
(n ≥ 3),
(5)
where 0.648214 ≈ ( 73 − eγ log log 12) log log 12 and the left-hand side of (5) attains its maximum at n = 12. In the same spirit, Lagarias [15] proved the equivalence of the RH to the
problem
(n ≥ 1),
σ(n) ≤ eHn log Hn + Hn ,
Pn
where Hn := j=1 1/j is the nth harmonic number.
Investigating upper and lower bounds for arithmetic functions, Landau [16, pp. 216–219]
obtained the following limits:
lim inf
n→∞
ϕ(n) log log n
= e−γ ,
n
lim sup
n→∞
ϕ(n)
= 1,
n
where ϕ(n) is the Euler totient function, which is defined as the number of positive integers
not exceeding n that are relatively prime to n. It can also be expressed as a product extended
over the distinct prime divisors of n [3, Theorem 2.4] by
Y
1
.
ϕ(n) = n
1−
p
p|n
Nicolas [18, 19] proved that if the RH is true, then we have for all k ≥ 2
Nk
> eγ ,
ϕ(Nk ) log log Nk
(6)
Q
where Nk = kj=1 pj and pj is the jth prime. On the other hand, if the RH is false, then for
infinitely many k, inequality (6) is true and for infinitely many k, inequality (6) is false.
Compared to numbers Nk which are the smallest integers that maximize n/ϕ(n), there
are integers which play this role for σ(n)/n and they are called superabundant numbers. A
positive integer n is said to be superabundant [2, 23] if
σ(n)
σ(m)
>
n
m
for all m < n.
3
(7)
We will use the symbol SA to denote the set of superabundant numbers and also as an
abbreviation for the term “superabundant”. Briggs [5] described a computational study
of the successive maxima of the relative sum-of-divisors function σ(n)/n. He also studied
the density of these numbers. Wójtowicz [26] showed that the values of the function f
defined in (2) are close to 0 on a set of asymptotic density 1. Another study on Robin’s
inequality (1) is due to Choie et al. [9]. They have shown that the RH holds true if and only
if every natural number divisible by a fifth power greater than 1 satisfies Robin’s inequality
(1).
Akbary and Friggstad [1] established the following interesting theorem which enables us
to limit our attention to a narrow sequence of positive integers, in order to find a probable
counterexample to inequality (1).
Theorem 2 ([1, Theorem 3]). If there is any counterexample to Robin’s inequality (1), then
the least such counterexample is a superabundant number.
Unfortunately, to our knowledge, there is no known algorithm (except the formula (7) in
the definition) to produce SA numbers. Alaoglu and Erdős [2] proved that
Q(x) > c
log x log log x
,
(log log log x)2
where Q(x) denotes the number of SA numbers not exceeding x. Later, Erdős and Nicolas
[10] demonstrated a stronger result that for every δ < 5/48 we have
Q(x) > (log x)1+δ ,
(x > x0 ).
As a natural question in this direction, it is interesting to introduce and study a set of
positive integers to which the first probable violation of inequality (1) belongs. Following this
aim, we introduce the sequence of extremely abundant numbers. We will establish another
criterion equivalent to the RH by proving that the RH is true if and only if there are
infinitely many extremely abundant numbers. Also, we give a connection between extremely
abundant numbers and CA numbers. Moreover, we present approximate formula for the
prime factorization of (sufficiently large) extremely abundant numbers. Finally, we establish
the fragility of the RH with respect to the terms of certain subsets of SA numbers which are
quite close to the set of extremely abundant numbers.
Before stating the main definition and results of this paper, we mention recent work of
Caveney et al. [6]. They defined a positive integer n as an extraordinary number if n is
composite and f (n) ≥ f (kn) for all
k ∈ N ∪ {1/p : p is a prime factor of n}.
Under these conditions they showed that the smallest extraordinary number is n = 4. Then
they proved that the RH is true if and only if 4 is the only extraordinary number. For more
properties of these numbers and comparisons with SA and CA numbers, we refer the reader
to [7].
4
2
Extremely abundant numbers
We define a new sequence of positive integers related to the RH. Our primary contribution
and motivation of this definition are Theorems 6 and 7. Let us now state the main definition
of this paper.
Definition 3. A positive integer n is extremely abundant if either n = 10080, or n > 10080
and
σ(n)
σ(m)
>
,
for all 10080 ≤ m < n.
(8)
n log log n
m log log m
Here 10080 has been chosen as the smallest SA number greater than 5040. In Table 1 we
list the first 20 extremely abundant numbers. To find them, we used a list of SA numbers
(see Proposition 5) provided by Kilminster [14] and Noe [20].
Remark 4. If we choose (instead of 10080) n1 such that 2520 < n1 ≤ 5040, and define n to
be extremely abundant if either n = n1 , or n > n1 and
σ(n)
σ(m)
>
,
n log log n
m log log m
for all n1 ≤ m < n,
then we have a finite number of elements n ≤ 5040 that satisfy the above inequality. Using
inequality (5), we have
f (n) < eγ +
0.648214
< f (5040),
(log log n)2
for some s10308 < n ≤ s10309 ,
where sk denotes the kth SA number listed in [20]. Checking by computer for values n
between 5040 and s10309 , we derive a finite set with maximum 5040. Similarly, one can easily
check for n1 < 2520, and get only sets with finite number of elements.
Let XA denote the set of all extremely abundant numbers. (We also use XA as an
abbreviation for the term “extremely abundant”.) Clearly, XA 6= CA (see Table 1). Indeed,
we shall prove that infinitely many elements of CA are not in XA and that, if RH holds,
then infinitely many elements of XA are in CA. As an elementary result from the definition
of XA numbers we have the following proposition.
Proposition 5. The inclusion XA ⊂ SA holds.
Proof. First, 10080 ∈ SA. Next, if n > 10080 and n ∈ XA, then for 10080 ≤ m < n we
have
σ(m)
σ(n)
= f (n) log log n > f (m) log log m =
.
n
m
In particular, for m = 10080 we get
σ(n)
σ(10080)
>
.
n
10080
5
So for m < 10080, we have
σ(n)
σ(10080)
σ(m)
>
>
n
10080
m
since 10080 ∈ SA. Therefore, n belongs to SA.
Next, motivating our construction of XA numbers, we will establish the first interesting
result of the paper.
Theorem 6. If there is any counterexample to Robin’s inequality (1), then the least one is
an XA number.
Proof. By doing some computer calculations we observe that there is no counterexample to
Robin’s inequality (1) for 5040 < n ≤ 10080. Now let n > 10080 be the least counterexample
to inequality (1). For m satisfying 10080 ≤ m < n we have
f (m) < eγ ≤ f (n).
Therefore n is an XA number.
As we mentioned in Section 1 we will prove an equivalent criterion to the RH for which
the proof is based on Robin’s inequality (1) and Gronwall’s theorem. Let #A denote the
cardinality of the set A. The second stimulus result is the following theorem. This result
also has its own interest that will be discussed in Section 5.
Theorem 7. The RH is true if and only if #XA = ∞.
Proof. Sufficiency. Assume that RH is not true. Then by Theorem 6 we have f (m) ≥ eγ
for some m ≥ 10080. From Gronwall’s theorem, we know that M = supn≥10080 f (n) is finite
and there exists n0 such that f (n0 ) = M ≥ eγ (if M = eγ then set n0 = m). An integer
n > n0 satisfies f (n) ≤ M = f (n0 ) and n can not be in XA, so #XA ≤ n0 .
Necessity. On the other hand, if RH is true, then Robin’s inequality (1) holds. If #XA
is finite, then let m be its largest element. For every n > m the inequality f (n) ≤ f (m)
holds and therefore
lim sup f (n) ≤ f (m) < eγ ,
n→∞
which is a contradiction to Gronwall’s theorem.
There are some primes which cannot be the largest prime factor of any XA number.
For example, referring to Table 1, there is no XA number with the largest prime factor
p(n) = 149. Do there exist infinitely many such primes?
6
3
Auxiliary lemmas and inequalities
Chebyshev’s functions ϑ(x) and ψ(x) are defined by
ϑ(x) =
X
log p,
X
ψ(x) =
log p =
pm ≤x
p≤x
X log x p≤x
log p
log p,
where ⌊x⌋ denotes the largest integer not exceeding x. It is known that the prime number
theorem (PNT) ([12, Theorem 434] and [13, Theorem 3, 12]) is equivalent to
ψ(x) ∼ x.
(9)
In his mémoir, Chebyshev proved the following lemma that we call Chebyshev’s result.
Lemma 8 ([8, p. 379]). For all x > 1
1
5
6
5
ϑ(x) < Ax − Ax 2 +
log2 x + log x + 2
5
4 log 6
2
15
12 1
5
log2 x −
log x − 3,
ϑ(x) > Ax − Ax 2 −
5
8 log 6
4
where
1
1
1
22 33 55
≈ 0.92129202.
1
30 30
We will use the following corollary in the proof of Theorem 26.
A = log
Corollary 9. We have
x
,
(x ≥ 3).
3
The next lemma here provides Littlewood’s result for oscillation of Chebyshev’s ϑ function.
ϑ(x) >
Lemma 10 ([7, Lemma 4]). There exists a constant c > 0 such that for infinitely many
primes p we have
√
ϑ(p) < p − c p log log log p,
(10)
and for infinitely many other primes p we have
√
ϑ(p) > p + c p log log log p.
In what follows, we shall frequently use the following elementary inequalities:
t
< log(1 + t) < t,
1+t
and
2t
< log(1 + t),
2+t
7
(t > 0),
(t > 0).
(11)
(12)
4
Some properties of SA, CA and XA numbers
We divide this section into three subsections, for which we shall exhibit several properties of
SA, CA and XA numbers, respectively. We denote by p(n) the largest prime factor of n or,
when there is no ambiguity, simply by p.
4.1
SA Numbers
Proposition 11. Let n < n′ be two consecutive SA numbers. Then
n′
≤ 2.
n
Proof. Let n = 2k2 · · · p. We compare n with 2n. In fact,
2k2 +2 − 1
σ(2n)/(2n)
= k2 +2
> 1.
σ(n)/n
2
−2
Hence n′ ≤ 2n.
Corollary 12. For any positive real number x ≥ 1 there exists at least one SA number n
such that x ≤ n < 2x.
Alaoglu and Erdős [2] have shown that if n = 2k2 · 3k3 · · · pkp is an SA number, then
k2 ≥ k3 ≥ · · · ≥ kp and kp = 1, except for n = 4, 36.
Proposition 13 ([2, Theorem 2]). Let n ∈ SA, and let q < r be prime factors of n with
corresponding exponents kq and kr . Set
kq log q
.
β :=
log r
Then kr has one of the three values: β − 1, β + 1, β.
As we observe, the above proposition determines the exponent of each prime factor of an
SA number with error of at most 1 in terms of a smaller prime factor of that number. In the
next lemma we prove a relation between the lower bound of an exponent of a prime factor
of n and its largest prime factor p.
Lemma 14. Let n ∈ SA, and let q be a prime factor of n. Then
log p
≤ kq .
log q
8
Proof. If q = p (= p(n)), then the result is trivial. Let q < p and kq = k. Suppose that
k ≤ ⌊log p/ log q⌋ − 1. Then
q k+1 < p.
(13)
Now we compare values of σ(ν)/ν, taking ν = n and ν = m = nq k+1 /p. Since σ(ν)/ν
is multiplicative, we restrict our attention to different factors. But n is an SA number
and m < n. Hence
1
σ(n)/n
q 2k+2 − q k+1
1
1
1+
1<
=
1+
=
.
σ(m)/m
q 2k+2 − 1
p
1 + 1/q k+1
p
Consequently, p < q k+1 , which contradicts inequality (13).
Proposition 15 ([2, Theorem 5]). Let n ∈ SA. If kq = k and q < (log p)α , where α is a
constant, then
q k+1 − 1
(log log p)2
log q
log k+1
1+O
,
(14)
>
q
−q
p log p
log p log q
q k+2 − 1
(log log p)2
log q
log k+2
1+O
.
(15)
<
q
−q
p log p
log p log q
Lemma 16. Let n ∈ SA, and let q be a fixed prime factor of n. Then there exist two positive
constants c and c ′ (depending on q) such that
cp
log p
log p
< q kq < c ′ p
.
log q
log q
Proof. By inequality (11)
q−1
q−1
1
q k+1 − 1
< k+1
= log 1 + k+1
≤ k
log k+1
q
−q
q
−q
q
−q
q
and (14), there exists a c ′ > 0 such that
qk < c ′
p log p
.
log q
On the other hand, again by inequality (11)
q−1
q−1
1
q k+2 − 1
> k+2
= log 1 + k+2
> k+1
log k+2
q
−q
q
−q
q
−1
2q
and (15), there exists a c > 0 such that
qk > c
p log p
.
log q
9
Corollary 17. Let n = 2k · · · p be an SA number. Then for n sufficiently large we have
k log 2
= 1.
log p
We showed [17] that for sufficiently large n ∈ SA
1
log n < p 1 +
.
2 log p
Our computation on the list of SA numbers in [20] suggests that a weaker inequality
2
log n < p 1 +
3 log p
(16)
(17)
holds for all n ≥ s365 . The product of exponent of a prime factor and the logarithm of the
corresponding prime factor of an SA number can be controlled, on average, by the logarithm
of the largest prime factor of that number. More precisely,
Proposition 18 ([2, Theorem 7]). If n ∈ SA, then
p(n) ∼ log n.
The next proposition gives a lower bound of n ∈ SA in terms of Chebyshev’s ψ function
compared to the above-mentioned asymptotic relation.
Proposition 19. Let n ∈ SA. Then
ψ(p(n)) ≤ log n.
Moreover,
lim
n→∞
n∈SA
ψ(p(n))
= 1.
log n
Proof. In fact, by Lemma 14
X
X log p(n) log q ≤
kq log q = log n.
ψ(p(n)) =
log q
q≤p(n)
q≤p(n)
To prove (18) we appeal to (the equivalent of) the PNT (9) and Proposition 18.
10
(18)
4.2
CA Numbers
By the definition of CA numbers (4) it is easily seen that CA ⊂ SA. Here we give a concise
description of the algorithm (essentially borrowed from [7, 10, 24]) to produce CA numbers.
For more details on this introduction we refer the reader to [2, 7, 10, 24].
Let F be defined by
F (x, k) =
log(1 + 1/(x + · · · + xk ))
.
log x
For ε > 0 we define x1 = x1 (ε) to be the only number such that
F (x1 , 1) = ε,
(19)
and xk = xk (ε) (for k > 1) to be the only number such that
F (xk , k) = ε.
Let
Ep = {F (p, α) : α ≥ 1}, p is a prime
and
E=
[
p
Ep = {ε1 , ε2 , . . .}.
If ε ∈
/ E, then the function σ(n)/n1+ε attains its maximum at a single point Nε whose prime
decomposition is
\$
%
p1+ε −1
Y
log
ε
p −1
Nε =
pαp (ε) ,
αp (ε) =
− 1,
log p
or if one prefers
αp (ε) =
(
k, if xk+1 < p < xk , k ≥ 1;
0, if p > x = x1 .
If ε ∈ E, then at most two xk ’s are prime. Hence, there are either two or four CA numbers
of parameter ε, defined by
K Y
Y
Nε =
p,
(20)
k=1 p<xk
or
p≤xk
where K is the largest integer such that xK ≥ 2. In particular, if N is the largest CA number
of parameter ε, then
F (p, 1) = ε ⇒ p(N ) = p,
(21)
where p(N ) is the largest prime factor of N . Therefore for any ε, formula (20) gives all
possible values of a CA number N of parameter ε [7].
11
Robin [24, Proposition 1] proved that the maximum order of the function f defined in (2)
is attained by CA numbers. More precisely, if 3 ≤ N < n < N ′ , where N and N ′ are two
successive CA numbers, then
f (n) ≤ max{f (N ), f (N ′ )}.
In the next proposition we improve the above inequality to a strict one.
Proposition 20. Let 3 ≤ N < n < N ′ , where N and N ′ are two successive CA numbers.
Then
f (n) < max{f (N ), f (N ′ )}.
(22)
Proof. In fact, due to the strict convexity of the function t 7→ εt − log log t, Robin’s proof
extends to the strict inequality (22).
Proposition 20 shows that if there is a counterexample to inequality (1), then there exists
at least one CA number that violates it.
Lemma 21. Let N < N ′ be two consecutive CA numbers. If there exists an XA number
n > 10080 satisfying N < n < N ′ , then N ′ is also an XA.
Proof. Let us set
B = {m ∈ XA : N < m < N ′ }.
By assumption n ∈ XA, we have B 6= ∅. Let n′ = max B. Since n′ ∈ XA and n′ >
N , it follows that f (n′ ) > f (N ). From inequality (22) we must have f (n′ ) < f (N ′ ).
Hence N ′ ∈ XA.
Remark 22. If n = 10080, then we have N = 5040 and N ′ = 55440, and
f (N ) ≈ 1.7909,
f (n) ≈ 1.7558,
f (N ′ ) ≈ 1.7512.
Hence, f (N ′ ) < f (n) < f (N ) and inequality (22) is satisfied with
f (n) < f (N ) = max{f (N ), f (N ′ )}.
Therefore N ′ ∈
/ XA. The point here is that n = 10080 is the initial XA number, so that
it misses the property (8) of the definition of XA numbers which is used in the proof of
Lemma 21.
Theorem 23. If RH holds, then there exist infinitely many CA numbers that are also XA.
Proof. If RH holds, then #XA = ∞ by Theorem 7. Let n ∈ XA. Since #CA = ∞ [2, 10],
there exist two successive CA numbers N, N ′ such that N < n ≤ N ′ . If N ′ = n then it is
already in XA, otherwise N ′ belongs to XA via Lemma 21.
It can be seen that there exist infinitely many CA numbers N for which the largest prime
factor p (= p(N )) is greater than log N . For this purpose we use the following lemma.
12
Lemma 24 ([7, Lemma 3]). Let N be a CA number of parameter ε with
ε < F (2, 1) = log(3/2)/ log 2
and define x = x(ε) by (19). Then
(i) for some constant c > 0
√
log N ≤ ϑ(x) + c x.
(ii) Moreover, if N is the largest CA number of parameter ε, then
√
ϑ(x) ≤ log N ≤ ϑ(x) + c x.
Theorem 25. There are infinitely many CA numbers Nε such that log Nε < p(Nε ).
Proof. Let p be sufficiently large satisfying the inequality (10), and let Nε be the largest CA
number of parameter
ε = F (p, 1).
Then from (21) it follows that p(Nε ) = p. By part (ii) of Lemma 24 we have
√
log Nε − ϑ(p) < c p,
(for some c > 0).
On the other hand, by Lemma 10 there exists a constant c′ > 0 such that
√
ϑ(p) − p < −c′ p log log log p,
Hence
(c′ > 0).
√
log Nε − p < {c − c′ log log log p} p < 0,
which is the desired conclusion.
4.3
XA Numbers
We return to XA numbers and present some of their properties. We begin by the first
interesting property of the XA numbers whose proof is essentially an application of the
definition of XA numbers.
Theorem 26. Let n ∈ XA. Then
p(n) < log n.
Proof. For n = 10080 we have
p(10080) = 7 < 9.218 < log(10080).
13
Let n > 10080 be an XA number and m = n/p(n). Then m > 10080, since for all primes p
we have ϑ(p) > p/3 (Corollary 9). Thus for n ∈ SA we have
1
log n ≥ ψ(p(n)) ≥ ϑ(p(n)) > p(n)
3
and m = n/p(n) > n/(3 log n) > 10080 if n ≥ 400, 000. For n < 400, 000 we can check by
computation. Hence by Definition 3 we obtain
1+
1
σ(n)/n
log log n
=
>
.
p(n)
σ(m)/m
log log m
Therefore,
log log n
log(1 + log p(n)/ log m)
1
>
−1=
.
p(n)
log log m
log log m
Using inequality (11) we have
1
log p(n)
log p(n)
>
>
⇒ p(n) < log n.
p(n)
log n log log m
log n log log n
We mention here a similar result proved by Choie et al. [9].
Proposition 27 ([9, Lemma 6.1]). Let t ≥ 2 be fixed. Suppose that there exists a t-free
integer exceeding 5040 that does not satisfy Robin’s inequality (1). Let n be the smallest such
integer. Then p(n) < log n.
In Theorem 23 we showed that if RH holds, then there exist infinitely many CA numbers
that are also XA. Next theorem is a conclusion of Theorems 25 and 26 which is independent
of the RH.
Theorem 28. There exist infinitely many CA numbers that are not XA.
We conclude this subsection with a result describing the structure of (sufficiently large)
XA numbers. More precisely, the next theorem will determine the exponents of the prime
factors of a (sufficiently large) XA number with an error at most 1.
Theorem 29. Let n = 2k2 · · · q kq · · · p ∈ XA. Set
p log p
.
αq (p) = logq 1 + (q − 1)
q log q
Then for sufficiently large n ∈ XA we have |kq − αq (p)| ≤ 1.
14
Proof. Assume that kq = k and k − αq (p) ≥ 2. Then we have
p log p
k
αq (p)+2
.
q ≥q
> q 1 + (q − 1)
q log q
(23)
Now let us compare f (n) with f (m), where m = n/q. Since n ∈ XA we must have
q k+1 − 1
log log n
σ(n)/n
= k+1
>
,
σ(m)/m
q
−q
log log m
or using inequality (11),
q k < 1 + (q − 1)
log n log log m
.
q log q
(24)
Comparison of (23) and (24) gives that
log n log log m − qp log p > q log q.
This contradicts inequality (16). Now we assume that k − αq (p) ≤ −2. Then
p log p
q k+2 − 1
≤
.
q−1
q log q
(25)
Choose m = nq/p. Comparing f (n) with f (m) we have
1
log log n
σ(n)/n
q−1
log p/q
1+
>
= 1 − k+2
>1+
.
σ(m)/m
q
−1
p
log log m
log n log log m
or simply by (25)
q log q
−
log p
1
p log p/q
− 1.
1+
>
p
log n log log m
Hence by (16) we have
q log q
−
log p
1
1+
p
which is false for all 2 ≤ q ≤ p.
>
(log p) 1 +
log p/q
1
1+
2 log p
1
2 log2 p
− 1,
Remark 30. Note that if inequality (17) holds for XA numbers n ≥ s365 , then by performing
computations for two smaller values of XA numbers, i.e., s20 and s356 (cf. Table 1) we see
that the above theorem holds true for all n ∈ XA.
15
5
Fragility of the RH and certain supersets of XA numbers
In Theorem 7 we proved that under the RH the cardinality of the set of XA numbers is
infinite. Here we present some interesting theorems which demonstrate the fragility of the
RH showing the infinitude of some supersets of XA numbers independent of the RH. These
sets are defined by inequalities quite close to that in (8). The basic inequalities used here to
define these sets are (11) and (12).
Lemma 31. If m ≥ 3, then there exists n > m such that
log n/m
σ(n)/n
>1+
.
σ(m)/m
log n log log m
Proof. Let m ≥ 3. Then by inequality (5)
σ(m)
0.648214
γ
log log m.
≤ e +
m
(log log m)2
Since for m′ > m
log log m
log log m′
1+
log m′ /m
log m′ log log m
(26)
<1
and the left-hand side is decreasing with respect to m′ and tends to zero as m′ → ∞, then
for some m′ > m we have
log log m
0.648214
log m′ /m
γ
e +
1+
= eγ − ε,
(27)
log log m′
log m′ log log m
(log log m)2
where ε > 0 is arbitrarily small and fixed. Hence by Gronwall’s theorem there exists n ≥ m′
such that
σ(n)
> (eγ − ε) log log n
n
log log m
log m′ /m
0.648214
γ
=
1+
log log n
e +
log log m′
log m′ log log m
(log log m)2
log n/m
σ(m)
≥ 1+
,
log n log log m
m
where the last inequality holds by (26) and (27).
Definition 32. Let n1 = 10080, and let nk+1 be the least integer greater than nk such that
log nk+1 /nk
σ(nk+1 )/nk+1
>1+
,
σ(nk )/nk
log nk+1 log log nk
We define X ′ to be the set of all n1 , n2 , n3 , . . . .
16
(k = 1, 2, . . .).
One can easily show that
XA ⊂ X ′ ⊂ SA.
(28)
Our first result towards the fragility of the RH is the following theorem.
Theorem 33. The set X ′ has an infinite number of elements.
Proof. If the RH is true, then the cardinality of X ′ is infinite by (28). If RH is not true,
then by Theorem 7 there exists m0 ≥ 10080 such that
log log m0
σ(m0 )/m0
≥
,
σ(m)/m
log log m
for all m ≥ 10080.
By Lemma 31 there exists m′ > m0 such that m′ satisfies
σ(m′ )/m′
log m′ /m0
>1+
.
σ(m0 )/m0
log m′ log log m0
Let n be the least number greater than m0 for which
log n/m0
σ(n)/n
>1+
.
σ(m0 )/m0
log n log log m0
Hence n ∈ X ′ .
The following lemma can be proved in the same manner as Lemma 31.
Lemma 34. If m ≥ 3, then there exists n > m such that
2 log n/m
σ(n)/n
>1+
.
σ(m)/m
(log m + log n) log log m
We continue our approach towards the fragility of the RH via (the stronger) inequality
(12) defining a smaller superset of XA numbers as follows.
Definition 35. Let n1 = 10080, and let nk+1 be the least integer greater than nk , such that
2 log nk+1 /nk
σ(nk+1 )/nk+1
>1+
,
σ(nk )/nk
(log nk + log nk+1 ) log log nk
(k = 1, 2, . . .).
We define X ′′ to be the set of all n1 , n2 , n3 , . . . .
By elementary inequality (12) and
t
2t
<
,
1+t
2+t
(t > 0)
one can easily show the inclusion XA ⊂ X ′′ ⊂ X ′ . The following theorem is a refinement of
Theorem 33 with a similar proof.
17
Theorem 36. The set X ′′ has an infinite number of elements.
We calculated the number of elements in XA, X ′ and X ′′ up to the 300, 000th element
of SA in [20]. Note that
#XA = 9240,
#X ′ = 9535,
#X ′′ = 9279
and
#(X ′ − XA) = 295,
#(X ′′ − XA) = 39.
It might be interesting to look at the list of elements of X ′′ − XA up to s300,000 :
X ′′ − XA = {s55 , s62 , s91 , s106 , s116 , s127 , s128 , s137 , s138 , s149 , s181 , s196 , s212 , s219 ,
s224 , s231 , s232 , s246 , s247 , s259 , s260 , s263 , s272 , s273 , s276 , s288 , s294 ,
s299 , s305 , s311 , s317 , s330 , s340 , s341 , s343 , s354 , s65343 , s271143 , s271151 }
Note that the second XA number is s356 (see Table 1) and only three out of 39 elements in
the set X ′′ − XA up to s300,000 , namely s65343 , s271143 and s271151 , are greater than s356 .
6
Numerical experiments
We present here some numerical results, mainly for the set of XA numbers (sorted in increasing order) up to 13770th element, which is less than C1 := s500,000 , based on the list
provided by Noe [20].
Property 37. Let n = 2k2 · · · q kq · · · rkr · · · p ∈ XA, where 2 ≤ q < r ≤ p. Then for
10080 < n ≤ C1
(i) log n < q kq +1 ,
(ii) rkr < q kq +1 < rkr +2 ,
(iii) q kq < kq p,
(iv) q kq log q < log n log log n < q kq +2 .
Property 38. Let n = 2k2 · · · xkk · · · p ∈ XA, where 2 < xk < p is the greatest prime factor
of exponent k. Then
p
√
p < x2 < 2p,
for 10080 < n ≤ C1 .
′
′
Property 39. Let n = 2k2 · · · q kq · · · p and n′ = 2k2 · · · q kq · · · p′ be two consecutive XA
numbers. Then for 10080 < n < n′ ≤ C1
|kq − kq′ | ≤ 1,
for all 2 ≤ q ≤ p′ .
18
Property 40. If m, n are XA numbers, then for 10080 ≤ m < n ≤ C1
(i) p(m) ≤ p(n),
(ii) d(m) ≤ d(n).
Remark 41. We note that Property 40 is not true for SA numbers. For instance,
s47 = (19♯)(3♯)2 2,
s48 = (17♯)(5♯)(3♯)23 ,
p(s47 ) = 19 > 17 = p(s48 ).
and
s173 = (59♯)(7♯)(5♯)(3♯)2 23 ,
s174 = (61♯)(7♯)(3♯)2 22 ,
d(s173 ) = 5308416 > 5160960 = d(s174 ).
Property 42. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C1
n′
(log log n)2
>1+c
,
n
log n
(log log n)2
n′
>1+c √
,
n
log n
(0 < c ≤ 4),
(0 < c ≤ 0.195).
Property 43. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C1
f (n′ )
1
< 1 + ′,
f (n)
p
where p ′ is the largest prime factor of n′ .
We have checked the following properties up to C2 = s250,000 and up to 8150th element
of XA numbers which is less than C2 .
Property 44. If n, n′ ∈ SA are consecutive, then
1
σ(n′ )/n′
< 1 + ′,
σ(n)/n
p
(n′ < C2 ),
where p ′ is the largest prime factor of n′ .
The number of distinct prime factors of a number n is denoted by ω(n) (see [25]).
Property 45. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C2 , then
n′
n
≤
.
ω(n)
ω(n′ )
19
The comparison of the sets CA and XA is given. We calculated them up to C = s1,000,000
from the list of SA in [20].
#{n ∈ XA : n < C} = 24875,
#{n ∈ CA : n < C} = 21187,
#{n ∈ CA ∩ XA : n < C} = 20468,
#{n ∈ XA \ CA : n < C} = 4407,
#{n ∈ CA \ XA : n < C} = 719,
We conclude this paper with another remark on choosing the first element of XA as 10080.
Remark 46. If we replace n1 = 10080, the initial number in the definition of XA numbers,
by n1 = 665280, we do not need to pose the condition (i.e., n > n1 ) in Lemma 21. Indeed, if
we choose the initial number n1 = 665280, then N = 55440 < 665280 < 720720 = N ′ , where
in this case N ′ is also an XA number. Therefore we do not need Remark 22. Moreover, if
we choose n1 = 665280, then there are 37 more XA numbers.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
n
s20 = (7♯)(3♯)23 = 10080
s356 = (113♯)(13♯)(5♯)(3♯)2 23
s368 = (127♯)(13♯)(5♯)(3♯)2 23
s380 = (131♯)(13♯)(5♯)(3♯)2 23
s394 = (137♯)(13♯)(5♯)(3♯)2 23
s408 = (139♯)(13♯)(5♯)(3♯)2 23
s409 = (139♯)(13♯)(5♯)(3♯)2 24
s438 = (151♯)(13♯)(5♯)(3♯)2 23
s440 = (151♯)(13♯)(5♯)(3♯)2 24
s444 = (151♯)(13♯)(7♯)(3♯)2 24
s455 = (157♯)(13♯)(5♯)(3♯)2 24
s458 = (157♯)(13♯)(7♯)(3♯)2 23
s459 = (157♯)(13♯)(7♯)(3♯)2 24
s476 = (163♯)(13♯)(7♯)(3♯)2 24
s486 = (163♯)(17♯)(7♯)(3♯)2 24
s493 = (167♯)(13♯)(7♯)(3♯)2 24
s502 = (167♯)(17♯)(7♯)(3♯)2 24
s519 = (173♯)(17♯)(7♯)(3♯)2 24
s537 = (179♯)(17♯)(7♯)(3♯)2 24
s555 = (181♯)(17♯)(7♯)(3♯)2 24
Type
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
f (n)
1.75581
1.75718
1.75737
1.75764
1.75778
1.75821
1.75826
1.75831
1.75849
1.75860
1.75864
1.75866
1.75892
1.75914
1.75918
1.75943
1.75966
1.76006
1.76038
1.76089
Table 1: First 20 extremely abundant numbers (pk ♯ :=
abundant number).
20
Qk
j=1
p(n)
7
113
127
131
137
139
139
151
151
151
157
157
157
163
163
167
167
173
179
181
log n
9.21831
126.444
131.288
136.163
141.083
146.018
146.711
156.039
156.732
158.678
161.788
163.041
163.734
168.828
171.661
173.946
176.779
181.933
187.120
192.318
k2
5
8
8
8
8
8
9
8
9
9
9
8
9
9
9
9
9
9
9
9
pj and c represents a colossally
7
Acknowledgments
We thank J. C. Lagarias, C. Calderon, J. Stopple, and M. Wolf for useful discussion and
sending us some relevant references. Our sincere thanks to J.-L. Nicolas for careful reading of
the manuscript, helpful comments, and worthwhile suggestions which improved the presentation of the paper. The work of the first author is supported by the Calouste Gulbenkian
Foundation, under Ph.D. grant number CB/C02/2009/32. Research partially funded by
the European Regional Development Fund through the programme COMPETE and by the
Portuguese Government through the FCT under the project PEst-C/MAT/UI0144/2011.
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21
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2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37, 11Y70, 11K31.
Keywords: extremely abundant number, superabundant number, colossally abundant number, Robin’s theorem, Chebyshev’s function.
22
(Concerned with sequences A004394, A004490, and A217867.)
Received April 1 2013; revised versions received August 16 2013; January 15 2014. Published
in Journal of Integer Sequences, February 7 2014.